Results 1 to 6 of 6

Math Help - Parametric Arc Length

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    3

    Angry Parametric Arc Length

    Alright I'm pretty lost here. I can usually deal with these types of problems fine, but with the given equations I don't seem to be getting anywhere.

    Given two parametric functions:

    x(t)=5e−05tsin(2t)
    y(t)=5e−05tcos(2t)

    0t6.

    Find the integral.
    Find the arc length, 0t6.
    Find the arc length, [0).

    Getting the derivative of each term and squaring it yields some nasty numbers and I'm running out of submissions.

    The equation to go with that I've been using is:

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,135
    Thanks
    1013
    Quote Originally Posted by apples12 View Post
    Alright I'm pretty lost here. I can usually deal with these types of problems fine, but with the given equations I don't seem to be getting anywhere.

    Given two parametric functions:

    x(t)=5e−05tsin(2t) are the bold expressions exponents?
    y(t)=5e−05tcos(2t)

    0t6.

    Find the integral.
    Find the arc length, 0t6.
    Find the arc length, [0).

    Getting the derivative of each term and squaring it yields some nasty numbers and I'm running out of submissions.

    The equation to go with that I've been using is:

    ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    3
    sorry my formatting got messed up in the posting

    x(t)=sin(2t)5e^(-0.5t)
    y(t)=cos(2t)5e^(-0.5t)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    x(t) = 5 e^{t/2} \sin 2t

    y(t) = 5 e^{t/2} \cos 2t



    x' = 5(1/2 e^{t/2} \sin 2t + 2e^{t/2} \cos 2t)

    x' = 5e^{t/2} (1/2 \sin 2t + 2\cos 2t )

    in the same way

    y' = 5e^{t/2} (1/2 \cos 2t - 2\sin 2t )


    ok

    x'^2 + y'^2 = 25e^t ( 1/4 \sin ^2 2t + 2\sin 2t \cos 2t + 4 \cos ^2 2t + 1/4 \cos ^2 2t - 2 \sin 2t \cos 2t + 4\sin ^2 2t )

    x'^2 + y'^2 = 25e^t (1/4 + 4 ) and this is easy to integrate
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2009
    Posts
    3
    this is wrong, the exponents are negative
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,135
    Thanks
    1013
    Quote Originally Posted by apples12 View Post
    this is wrong, the exponents are negative
    so? ... take Amer's post and fix the signs.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parametric Arc Length
    Posted in the Calculus Forum
    Replies: 20
    Last Post: August 11th 2010, 01:16 PM
  2. parametric arc length
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 10th 2009, 01:12 PM
  3. Length of Curve (Parametric)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 26th 2009, 05:09 AM
  4. arc length parametric
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 10th 2008, 10:54 PM
  5. Arc length from parametric
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 20th 2008, 10:03 AM

Search Tags


/mathhelpforum @mathhelpforum